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Quadratic polynomials in practice

Solving exercises about roots of quadratic polynomials
Hello. Let us consider the first exercise of the quadratic polynomials in practice step. We want to write both these two quadratic polynomials in this form. Let us start with the first one. Then the first thing that we have to do is just to collect the 2, and we get 2 times x squared plus 2x minus 15. And now we try to write this quadratic polynomial as x plus b times x plus q. How can I do it? x squared plus 2x minus 15– the idea is to consider these two terms as the beginning of the quadratic of a binomial. You see, this is the square of x plus 2 times x times 1.
Now I complete the expansion of the quadratic of this binomial, adding 1 and subtracting 1. In such a way, I have again the same beginning, and then minus 15. Now you see, this is equal to what?
The first part is exactly x plus 1 squared, and then you have minus 16.
Now you have the difference between two squares, and therefore, this is equal exactly to x plus 1 minus 4, times x plus 1, plus 4, which is x minus 3 times x plus 5. Therefore, we get that our initial polynomial– 2 x squared plus 4x minus 30– is equal to 2 times x minus 3 times x plus 5. And now consider the other polynomial. The idea is exactly the same. 3 x squared plus 7x minus 6 is equal to 3 times x squared plus 7 over 3 x minus 2. And now this polynomial can be written in such a way again. This can be considered as the initial part of the expansion of the square of a binomial.
Indeed, this is equal to x squared plus– when you have the expansion of the quadratic of a binomial, you have the square, the first term. Then you have two times the first term by the second term. And for this reason, let me write this in such a way. 2 times 7/6 times x plus– now you’ll see 7/6 to the square and minus 7/6 to the square. You see, this gives us no contribution. And then we have minus 2. You see what I have realized. This part is the square of a binomial. Indeed, you have the square of x plus the square of 7/6 plus 2 times 7/6 times x. Then this is equal to what?
To x plus 7/6 squared minus this quantity, which is minus 49/36 minus 2, which is x plus 7/6 squared minus–
and I have a 36 here, 49 here, plus 72, which is equal to x plus 7/6 squared minus– and we have 72 plus 50 would be 122, plus 49 will be 121, over 36. And here again, you have the difference between two squares. This is then equal to x plus 7/6 plus the root square– the square root of this fraction, which is 11/6, times x minus– x plus 7/6 minus 11/6.
Therefore, we get that our polynomial– 3 x squared plus 7 times 6 minus 6– is equal to 3 times x plus 18/6, which is x plus 3, times x plus 7/6 and minus 11/6. 7/6 minus 11/6 is minus 4/6, which is minus 2/3.
And we have got our final result. Ciao.

The following exercises are solved in this step.

We invite you to try to solve them before watching the video.

In any case, you will find below a PDF file with the solutions.

Exercise 1.

Write the following polynomials in the form (a(x+p)(x+q)):

i) (2x^2+4x-30);

ii) (3x^2+7x-6).

Exercise 2. [Solved only in the PDF file]

Compute the roots of the polynomial (4x^2+7x-12).

Exercise 3. [Solved only in the PDF file]

Do the graphs of (f) and (g) intersect, if

i) (f:xmapsto x^2-3x+7) and (g:xmapsto 2x-3)?

ii) (f:xmapsto x^2-3x+7) and (g:xmapsto 2x+13)?

Exercise 4. [Solved only in the PDF file]

Determine the maximum value of (-2x^2+3x-7).

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Precalculus: the Mathematics of Numbers, Functions and Equations

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